fork from aQuaYi/LeetCode-in-Go

Author: ruoru

.travis.yml

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+language: rust
+
+sudo: required
+# e6f235e5-d272-40ee-87f0-60264eb6ce19
+rust:
+  - stable
+  - beta
+  - nightly
+
+matrix:
+  allow_failures:
+    - rust: nightly
+
+addons:
+  apt:
+    packages:
+      - libcurl4-openssl-dev
+      - libelf-dev
+      - libdw-dev
+      - cmake
+      - gcc
+      - binutils-dev
+      - libiberty-dev
+
+after_success: |
+  wget https://github.com/SimonKagstrom/kcov/archive/master.tar.gz &&
+  tar xzf master.tar.gz &&
+  cd kcov-master &&
+  mkdir build &&
+  cd build &&
+  cmake .. &&
+  make &&
+  make install DESTDIR=../../kcov-build &&
+  cd ../.. &&
+  rm -rf kcov-master &&
+  for file in target/debug/examplerust-*[^\.d]; do mkdir -p "target/cov/$(basename $file)"; ./kcov-build/usr/local/bin/kcov --exclude-pattern=/.cargo,/usr/lib --verify "target/cov/$(basename $file)" "$file"; done &&
+  bash <(curl -s https://codecov.io/bash) &&
+  echo "Uploaded code coverage"

Algorithms/0001.two-sum/README.md

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+# [1. Two Sum](https://leetcode.com/problems/two-sum/)
+
+## 题目
+
+Given an array of integers, return indices of the two numbers such that they add up to a specific target.
+
+You may assume that each input would have exactly one solution, and you may not use the same element twice.
+
+Example:
+
+```text
+Given nums = [2, 7, 11, 15], target = 9,
+
+Because nums[0] + nums[1] = 2 + 7 = 9,
+return [0, 1].
+```
+
+## 解题思路
+
+```rs
+a + b = target
+```
+
+也可以看成是
+
+```rs
+a = target - b
+```
+
+在`map[整数]整数的序号`中，可以查询到a的序号。这样就不用嵌套两个for循环了。

Algorithms/0001.two-sum/two-sum.rs

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+# [2. Add Two Numbers](https://leetcode.com/problems/add-two-numbers/)
+
+## 题目
+
+You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
+
+You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+
+```text
+Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
+Output: 7 -> 0 -> 8
+```
+
+## 解题思路
+
+```text
+(2 -> 4 -> 3)是 342
+
+(5 -> 6 -> 4)是 465
+
+(7 -> 0 -> 8)是 807
+
+342 + 465 = 807
+```
+
+所以，题目的本意是，把整数换了一种表达方式后，实现其加法。
+
+设计程序时候，需要处理的点有
+
+1. 位上的加法，需要处理进位问题
+1. 如何进入下一位运算
+1. 按位相加结束后，也还需要处理进位问题。
+
+## 总结
+
+读懂题意后，按步骤实现题目要求。

Algorithms/0001.two-sum/two-sum_test.rs

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+package problem0002
+
+import (
+	"github.com/aQuaYi/LeetCode-in-Go/kit"
+)
+
+// ListNode defines for singly-linked list.
+//  type ListNode struct {
+//      Val int
+//      Next *ListNode
+//  }
+type ListNode = kit.ListNode
+
+func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
+	resPre := &ListNode{}
+	cur := resPre
+	carry := 0
+
+	for l1 != nil || l2 != nil || carry > 0 {
+		sum := carry
+
+		if l1 != nil {
+			sum += l1.Val
+			l1 = l1.Next
+		}
+
+		if l2 != nil {
+			sum += l2.Val
+			l2 = l2.Next
+		}
+
+		carry = sum / 10
+
+		cur.Next = &ListNode{Val: sum % 10}
+		cur = cur.Next
+	}
+
+	return resPre.Next
+}

Algorithms/0002.add-two-numbers/README.md

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+package problem0002
+
+import (
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+type para struct {
+	one *ListNode
+	two *ListNode
+}
+
+type ans struct {
+	one *ListNode
+}
+
+type question struct {
+	p para
+	a ans
+}
+
+func makeListNode(is []int) *ListNode {
+	if len(is) == 0 {
+		return nil
+	}
+
+	res := &ListNode{
+		Val: is[0],
+	}
+	temp := res
+
+	for i := 1; i < len(is); i++ {
+		temp.Next = &ListNode{Val: is[i]}
+		temp = temp.Next
+	}
+
+	return res
+}
+
+func Test_OK(t *testing.T) {
+	ast := assert.New(t)
+
+	qs := []question{
+		question{
+			p: para{
+				one: makeListNode([]int{2, 4, 3}),
+				two: makeListNode([]int{5, 6, 4}),
+			},
+			a: ans{
+				one: makeListNode([]int{7, 0, 8}),
+			},
+		},
+		question{
+			p: para{
+				one: makeListNode([]int{9, 8, 7, 6, 5}),
+				two: makeListNode([]int{1, 1, 2, 3, 4}),
+			},
+			a: ans{
+				one: makeListNode([]int{0, 0, 0, 0, 0, 1}),
+			},
+		},
+		question{
+			p: para{
+				one: makeListNode([]int{0}),
+				two: makeListNode([]int{5, 6, 4}),
+			},
+			a: ans{
+				one: makeListNode([]int{5, 6, 4}),
+			},
+		},
+	}
+
+	for _, q := range qs {
+		a, p := q.a, q.p
+		ast.Equal(a.one, addTwoNumbers(p.one, p.two), "输入:%v", p)
+	}
+}

Algorithms/0002.add-two-numbers/add-two-numbers.rs

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+# [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/)
+
+## 题目
+Given a string, find the length of the longest substring without repeating characters.
+
+Examples:
+
+Given "abcabcbb", the answer is "abc", which the length is 3.
+
+Given "bbbbb", the answer is "b", with the length of 1.
+
+Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
+
+## 解题思路
+利用s[left:i+1]来表示s[:i+1]中的包含s[i]的最长子字符串。
+location[s[i]]是字符s[i]在s[:i+1]中倒数第二次出现的序列号。
+当left < location[s[i]]的时候，说明字符s[i]出现了两次。需要设置 left = location[s[i]] + 1,
+保证字符s[i]只出现一次。
+
+## 总结
+利用Location保存字符上次出现的序列号，避免了查询工作。location和[Two Sum](./algorithms/0001.two-sum)中的m是一样的作用。
+```go
+// m 负责保存map[整数]整数的序列号
+	m := make(map[int]int, len(nums))
+```

Algorithms/0002.add-two-numbers/add-two-numbers_test.rs

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+package problem0003
+
+func lengthOfLongestSubstring(s string) int {
+	// location[s[i]] == j 表示：
+	// s中第i个字符串，上次出现在s的j位置，所以，在s[j+1:i]中没有s[i]
+	// location[s[i]] == -1 表示： s[i] 在s中第一次出现
+	location := [256]int{} // 只有256长是因为，假定输入的字符串只有ASCII字符
+	for i := range location {
+		location[i] = -1 // 先设置所有的字符都没有见过
+	}
+
+	maxLen, left := 0, 0
+
+	for i := 0; i < len(s); i++ {
+		// 说明s[i]已经在s[left:i+1]中重复了
+		// 并且s[i]上次出现的位置在location[s[i]]
+		if location[s[i]] >= left {
+			left = location[s[i]] + 1 // 在s[left:i+1]中去除s[i]字符及其之前的部分
+		} else if i+1-left > maxLen {
+			// fmt.Println(s[left:i+1])
+			maxLen = i + 1 - left
+		}
+		location[s[i]] = i
+	}
+
+	return maxLen
+}

Algorithms/0003.longest-substring-without-repeating-characters/README.md

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+package problem0003
+
+import (
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+type para struct {
+	one string
+}
+
+type ans struct {
+	one int
+}
+
+type question struct {
+	p para
+	a ans
+}
+
+func Test_OK(t *testing.T) {
+	ast := assert.New(t)
+
+	qs := []question{
+		question{
+			p: para{
+				one: "abcabcbb",
+			},
+			a: ans{
+				one: 3,
+			},
+		},
+		question{
+			p: para{
+				one: "bbbbbbbb",
+			},
+			a: ans{
+				one: 1,
+			},
+		},
+		question{
+			p: para{
+				one: "pwwkew",
+			},
+			a: ans{
+				one: 3,
+			},
+		},
+	}
+
+	for _, q := range qs {
+		a, p := q.a, q.p
+		ast.Equal(a.one, lengthOfLongestSubstring(p.one), "输入:%v", p)
+	}
+}

Algorithms/0003.longest-substring-without-repeating-characters/longest-substring-without-repeating-characters.rs

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+# [4. Median of Two Sorted Arrays](https://leetcode.com/problems/median-of-two-sorted-arrays/)
+
+## 题目
+There are two sorted arrays nums1 and nums2 of size m and n respectively.
+
+Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
+
+Example 1:
+```
+nums1 = [1, 3]
+nums2 = [2]
+The median is 2.0
+```
+Example 2:
+```
+nums1 = [1, 2]
+nums2 = [3, 4]
+The median is (2 + 3)/2 = 2.5
+```
+## 解题思路
+输入两个已经排序好的整数切片，合并两个切片，返回合并后切片的中位数。
+
+所以，题目的关键是，如何高效地合并切片。
+
+## 总结
+分步骤完成程序
+